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2 years ago

What are the three major tendons in the knee?

Physics

2 answers:

kati45 [8]2 years ago

6 0

Answer:

The two important tendons in the knee are (1) the quadriceps tendon connecting the quadriceps muscle, which lies on the front of the thigh, to the patella and (2) the patellar tendon connecting the patella to the tibia (technically, this is a ligament because it connects two bones).

Fynjy0 [20]2 years ago

3 0

Three major tendons in the knee are the quadriceps tendon patellar tendon and the hamstring tendon (not sure about the last one but I think its that) Hope this helps.

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A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai

shutvik [7]

Answer:

d) None of the above

Explanation:

$v_{o}$ = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m

t = time of travel

using the equation

$y=v_{oy} t+(0.5)a_{y} t^{2}$

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

$v_{ox}$ = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

$a_{x}$ = acceleration along the horizontal direction = 0 m/s²

t = time of travel = 2.1 s

Using the kinematics equation

$x =v_{ox} t+(0.5)a_{x} t^{2}$

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

$y_{o}$ =initial vertical position at the time of launch = 20 m

$y$ = vertical position at the maximum height = 20 m

$v_{fy}$ = final velocity along vertical direction at highest point = 0 m/s

using the equation

${v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})$

$0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)$

$y$ = 20.02 m

h = height above the starting height

h = $y$ - $y_{o}$

h = 20.02 - 20

h = 0.02 m

7 0

3 years ago

Once an object enters orbit, what keeps the object moving sideways?

yaroslaw [1]

It's the natural tendency of things to keep going unless there's something trying to stop them.

It's usually called "inertia".

Don't get the idea from all of this that things stop unless there's something to keep them going. The truth is exactly the opposite: Things keep going unless there's something to make them stop.

7 0

3 years ago

Read 2 more answers

A copper rod of length 27.5 m has its temperature increases by 35.9 degrees celsius. how much does its length increase?(unit=m)

gavmur [86]

<h2>The increase in length = 1.87 x 10⁻²</h2>

Explanation:

When copper rod is heated , its length increases

The increase in length can be found by the relation

L = L₀ ( 1 + α ΔT )

here L is the increased length and L₀ is the original length

α is the coefficient of linear expansion and ΔT is the increase in temperature .